[SciPy-user] FFT speed ?

[Samuel GARCIA]

Yes I know this.
But 186888 186889 and 186890 are not power of 2 and the computation time
is very very different just for a difference of size of only one point.
What is the reason ?
And how to deal with that ? (I realy need to compute fft with a random
nomber of point)


Nils Wagner wrote:
> Samuel GARCIA wrote:
>   
>> Hello list,
>>
>> I have a problem with fft speed.
>> could someone try this :
>>
>> from scipy import *
>> import time
>>
>> a = rand((186888));t1=time.time();fft(a);t2=time.time();t2-t1
>>
>> # for me about 0.67791390419006348 s. on centrino 1.6GHz
>>
>> a = rand((186890));t1=time.time();fft(a);t2=time.time();t2-t1
>>
>> # for me about 1.622053861618042 s. on centrino 1.6GHz
>>
>> a = rand((186889));t1=time.time();fft(a);t2=time.time();t2-t1
>>
>> # the computation is infinit !!!! (no time to wait)
>>
>>
>> Question : why is there a problem specialy for that size (186889) of fft ?
>> It is not the only size with that problem.
>>
>>
>> Thank
>>
>> Sam
>>
>>
>>   
>>     
> What you can find with help (fft) is ...
>
> "This is most efficient for n a power of two"
>
> Nils
>
>
>  
>
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-- 
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Samuel Garcia
Universite Claude Bernard LYON 1
CNRS - UMR5020, Laboratoire des Neurosciences et Systemes Sensoriels
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FRANCE
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